# Uniform Circular Motion

Reference > Science > Physics > Study Guide > Unit 4: Kinematics 2 - Motion in Curved LinesThere are many situations in nature where an object moves around a circular path at a constant speed. The tip of the blade of a fan or airplane propellor, a fly riding on a longplaying record, or the stone in David's slingshot are all moving with uniform circular motion. The earth's motion around the sun and the moon's motion around the earth are both good approximations to uniform circular motion as are the motions of many communications satellites.

It is very important to recognize that for an object in uniform circular motion, the speed is constant but not the velocity. the velocity is not constant because the direction of motion is constantly changing. Since the velocity is changing the object must have an acceleration if we define acceleration as change in velocity (not speed) divided by the time. It can be shown that the following equation gives the magnitude of the acceleration for an object in uniform circular motion:_{c}=

^{2}

The equation above gives only the magnitude of the acceleration. The direction of the acceleration vector is always toward the center of the circle. Whenever working with uniform circular motion it is essential that the directions of the velocity and acceleration vectors be kept in mind. The velocity vector is always tangent to the circle and the acceleration vector is always radially inward. The acceleration is often called the central or centripetal acceleration.

Two other quantities are often used in connection with motion in a circle. They are also used with any periodic motion so you will use them often. Therefore be sure you learn their definitions well. The first is the period of the motion. The period is the amount of time required for one complete revolution in the case of circular motion, or one complete cycle for other cyclic motion. Since it is really a time, its units will be time units. The other is the frequency which is defined as the number of revolutions or cycles completed in a unit time, often cycles. Since it is a number/time, its units are the units of time^{-1}. The most common unit of frequency, often used in wave studies, is the Hertz which is 1/sec. The period and the frequency of a motion are reciprocals of each other, ie T =

Since T = 1/f and f= 1/T the previous equation can be written as:

_{c}= 4p

^{2}f

^{2}R

_{c}=

^{2}R

^{2}

Since an object moving in a circular path is being constantly accelerated it must have a resultant force acting on it. The force is called the centripetal force and its direction is toward the center of the circle. The magnitude of the force can be found by using Newton's 2nd law of motion, F = ma, and one of the equations above to calculate the acceleration.

## Sample Problem #1

A record, 10" in diameter is rotating at 33 1/3 rpm (revolutions per minute) on a turntable. There are two spots on the record, one is 2" from the center and the other is 4" out.

- Calculate the frequency and period of the record.
- Calculate the speed of each spot.
- Calculate the acceleration of each spot.

## Sample Solution #1

1. The frequency is given, 33 1/3 rpm. In physics we usually would use seconds and would write it differently. Divide 33 1/3 by 60 to find how many revolutions or cycles are made in one second and write the answer this way:

^{â1}

V = 2pR/T = 2p4/1.8 = 14.

3. To find the acceleration either acceleration equation can be used since we have the speed, frequency and period of each spot.

a_{c}= V

^{2}/R = (7.0)2/2 = 24.5

^{2}

a

_{c}= V

^{2}/R = (14.0)2/4 = 49

^{2}

## Sample Problem #2

David has a giant to kill and his only weapon is a sling. It consists of two leather thongs 36" long attached to a leather pouch in which David places a small stone. He spins it in a circular motion which has a radius of 40" making 40 revolutions in 10 seconds and then releases one leather thong sending the projectile toward the giant. Calculate the speed and central acceleration of the stone in the slingshot.

## Sample Solution #2

^{-1}

_{c})

V = 1005

_{c}= V

^{2}/R = (1005)2/40

a

_{c}= 25250

^{2}

## Sample Problem #3

The coefficient of friction between a car and the road is 0.25. Calculate the maximum speed at which a 1500 kg car can travel around a curve of radius 100m.

## Sample Solution #3

At the maximum possible speed, the force of friction is just sufficient to provide the centripetal acceleration. First we must calculate the possible friction force.

_{f}= μF

_{N}

F

_{N}= W = ma

_{g}= 1500(9.8)

F

_{N}= 14700N

Therefore

_{f}= 0.25(14700)

F

_{f}= 3675N

3675 = 1500a

a =

a = 2.45

^{2}

_{c}=

^{2}

2.45 =

^{2}

V = 15.6

Actually it is not necessary to give the mass of the car. It turns out that the mass of the car will cancel out during the calculations. Redo the problem using M for the mass of the car and you will see that the maximum speed is the same regardless of the mass of the vehicle as long as the coefficient of friction is the same.

## Sample Problem #4

A boy swings a pail full of water in a vertical circle so that at the top of the swing it is upside down. If the radius of the circle is 1 m, calculate the minimum number of swings the pail must make per minute if he is to stay dry.

## Sample Solution #4

When the pail is at the top of its swing the water has an acceleration of^{2}

^{2}

^{2}

^{2}

^{2}

^{2}

^{2}Rf

^{2}

9.8 = 4p

^{2}(1)f

^{2}

f = 0.5p s

^{-1}

If the frequency is 0.5p s than it must be 60 times 0.5 or 30p min^{-1}.

## Sample Problem #5

Roadbeds are frequently tilted or banked on curves so that the road will apply a force toward the center of the curve to help provide the centripetal force necessary to hold a car in a circular path. Calculate the angle of bank for a roadbed with a radius of 100m if a car is to go around the curve with no friction force.

## Sample Solution #5

Examine carefully the following diagram (Figure 4.2.1) which shows a cross section of the road with a car on it and the forces acting on the car. If there is to be no friction, the only two forces on the car will be the weight of the car (W) and the normal force of the road against the car (N) as shown. We need to have the resultant force toward the center of the circle so we will resolve the vectors into vertical and horizontal components as shown.

^{Figure 4.2.1}

Resolve the normal force into vertical and horizontal components.

_{x}= NsinΘ

N

_{y}= NcosΘ

The car does not accelerate in the vertical direction so

_{y}= W = ma

_{g}

_{x}=

^{2}

_{x}= NsinΘ =

^{2}

N

_{y}= NcosΘ = ma

_{g}

^{2}

_{g}

^{2}

_{g}

tanΘ =

^{2}

_{g}